A vertical tower stands on a horizontal ground. Two points P and Q...

Let the height of the tower be $h$ and its base be $O$.

The distance of point $P$ from the base of the tower is $OP = h \cot 45^\circ = h$.

The distance of point $Q$ from the base of the tower is $OQ = h \cot 30^\circ = h\sqrt{3}$.

Since $P$ is due East and $Q$ is due North of the tower, the angle $\angle POQ$ is $90^\circ$.

In the right-angled triangle $POQ$, applying the Pythagoras theorem:
$PQ^2 = OP^2 + OQ^2$

Substitute the given distance $PQ = 40$ m:
$40^2 = h^2 + (h\sqrt{3})^2$
$1600 = h^2 + 3h^2$
$1600 = 4h^2$
$h^2 = 400$
$h = 20$

Thus, the height of the tower is $20$ meters.

Answer: $20$